The number of ordered pairs $(x, 1)$ satisfying the equations $\sin x+\sin y=\sin (x+y)$ and $|x|+|y|=1$ is
The number of ordered pairs $(x, 1)$ satisfying the equations $\sin x+\sin y=\sin (x+y)$ and $|x|+|y|=1$ is
$2$
$3$
$4$
$6$
Solution
$\sin x+\sin y=\sin (x+y)$
$2 \sin \frac{(x+y)}{2} \cos \frac{(x-y)}{2}=2 \sin \frac{(x+y)}{2} \cos \frac{(x+y)}{2}$
$\Rightarrow \sin \frac{(x+y)}{2}\left[\cos \frac{(x-y)}{2}-\cos \frac{(x+y)}{2}\right]=0$
$\Rightarrow \sin \frac{(x+y)}{2} \sin \frac{x}{2} \sin \frac{y}{2}=0$
$\therefore$ Either $\sin \frac{x+y}{2}=0$ or $\sin \frac{x}{2}=0$ or $\sin \frac{y}{2}=0$
$\Rightarrow x+y=0$ or $x=0$ or $y=0$
Also $|x|+|y|=1$
$\Rightarrow x+y=1, \quad x-y=1$
$x+y=-1, \quad x-y=-1$
Thus solving $x+y=0$ with $x-y=1$ or $x-y=-1$
We get $\left(\frac{1}{2}, \frac{-1}{2}\right)$ or $\left(-\frac{1}{2}, \frac{1}{2}\right)$
Solving with $x=0$, we get $(0, \pm 1)$
Solving with $y=0$ we get $( \pm 1,0)$
So, we get 6 orderd pairs.